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Op-Amp Circuits • the Inverting Amplifier • the Non-Inverting Amplifier

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Op-Amp Circuits • the Inverting Amplifier • the Non-Inverting Amplifier Electronic Instrumentation Experiment 4 * Part A: Introduction to Operational Amplifiers * Part B: Voltage Followers * Part C: Integrators and Differentiators * Part D: Amplifying the Strain Gauge Signal Part A Introduction to Operational Amplifiers Operational Amplifiers Op-Amp Circuits • The Inverting Amplifier • The Non-Inverting Amplifier Operational Amplifiers Op-Amps are possibly the most versatile linear integrated circuits used in analog electronics. The Op-Amp is not strictly an element; it contains elements, such as resistors and transistors. However, it is a basic building block, just like R, L, and C. We treat this complex circuit as a black box. The Op-Amp Chip The op-amp is a chip, a small black box with 8 connectors or pins (only 5 are usually used). The pins in any chip are numbered from 1 (starting at the upper left of the indent or dot) around in a U to the highest pin (in this case 8). 741 Op Amp or LM351 Op Amp Op-Amp Input and Output The op-amp has two inputs, an inverting input (-) and a non-inverting input (+), and one output. The output goes positive when the non-inverting input (+) goes more positive than the inverting (-) input, and vice versa. The symbols + and – do not mean that that you have to keep one positive with respect to the other; they tell you the relative phase of the output. (Vin=V1-V2) A fraction of a millivolt between the input terminals will swing the output over its full range. Powering the Op-Amp Since op-amps are used as amplifiers, they need an external source of (constant DC) power. Typically, this source will supply +15V at +V and -15V at -V. We will use ±9V. The op-amp will output a voltage range of of somewhat less because of internal losses. The power supplied determines the output range of the op-amp. It can never output more than you put in. Here the maximum range is about 28 volts. We will use ±9V for the supply, so the maximum output range is about 16V. Op-Amp Intrinsic Gain Amplifiers increase the magnitude of a signal by multiplier called a gain -- “A”. The internal gain of an op-amp is very high. The exact gain is often unpredictable. We call this gain the open-loop gain or intrinsic gain. Vout 56 =Aopen loop ≈− 10 10 Vin The output of the op-amp is this gain multiplied by the input V out= Aol ⋅Vin = Aol (V1 −V2 ) Op-Amp Saturation The huge gain causes the output to change dramatically when (V1-V2) changes sign. However, the op-amp output is limited by the voltage that you provide to it. When the op-amp is at the maximum or minimum extreme, it is said to be saturated. −V ≤ Vout ≤ +V if V1 > V2 then Vout ≈ +V positive saturation if V1 < V2 then Vout ≈ −V negative saturation How can we keep it from saturating? Feedback Negative Feedback • As information is fed back, the output becomes more stable. Output tends to stay in the “linear” range. The linear range is when Vout=A(V1-V2) vs. being in saturation. • Examples: cruise control, heating/cooling systems Positive Feedback • As information is fed back, the output destabilizes. The op-amp tends to saturate. • Examples: Guitar feedback, stock market crash • Positive feedback was used before high gain circuits became available. Op-Amp Circuits use Negative Feedback Negative feedback couples the output back in such a way as to cancel some of the input. Amplifiers with negative feedback depend less and less on the open-loop gain and finally depend only on the properties of the values of the components in the feedback network. The system gives up excessive gain to improve predictability and reliability. Op-Amp Circuits Op-Amps circuits can perform mathematical operations on input signals: • addition and subtraction • multiplication and division • differentiation and integration Other common uses include: • Impedance buffering • Active filters • Active controllers • Analog-digital interfacing Typical Op Amp Circuit • V+ and V- power the op-amp • Vin is the input voltage signal • R2 is the feedback impedance R2 10k • R1 is the input impedance V+ 9Vdc • Rload is the load U2 7 3 5 0 V + V+ OS2 6 0 OUT R1 Vout 2 1 - 4 OS1 Vin 1k VOFF = 0 uA741 Rload V V- VAMPL = .2 1k FREQ = 1k V- -9Vdc 0 0 0 The Inverting Amplifier R f R f Vout = − Vin A = − Rin Rin The Non-Inverting Amplifier R = + f Vout 1 Vin Rg R A =1+ f Rg Remember to disconnect the batteries. End of part A Part B The Voltage Follower Op-Amp Analysis Voltage Followers Op-Amp Analysis We assume we have an ideal op-amp: • infinite input impedance (no current at inputs) • zero output impedance (no internal voltage losses) • infinite intrinsic gain • instantaneous time response Golden Rules of Op-Amp Analysis Rule 1: VA = VB • The output attempts to do whatever is necessary to make the voltage difference between the inputs zero. • The op-amp “looks” at its input terminals and swings its output terminal around so that the external feedback network brings the input differential to zero. Rule 2: IA = IB = 0 • The inputs draw no current • The inputs are connected to what is essentially an open circuit Steps in Analyzing Op-Amp Circuits 1) Remove the op-amp from the circuit and draw two circuits (one for the + and one for the – input terminals of the op amp). 2) Write equations for the two circuits. 3) Simplify the equations using the rules for op amp analysis and solve for Vout/Vin Why can the op-amp be removed from the circuit? • There is no input current, so the connections at the inputs are open circuits. • The output acts like a new source. We can replace it by a source with a voltage equal to Vout. Analyzing the Inverting Amplifier 1) inverting input (-): non-inverting input (+): How to handle two voltage sources VRin = Vin −VB VRf = VB −Vout R V = (V −V ) in Rin in out + R f Rin R V = (V −V ) f Rf in out + R f Rin 3k V = (5V − 3V ) =1.5V V = V +V = 4.5V Rf 3k +1k B out Rf Inverting Amplifier Analysis 1) − : + : V V −V V −V 2) − : i = = in B = B out R Rin R f + : VA = 0 Vin −Vout 3) VA = VB = 0 = Rin R f V R out = − f Vin Rin Analysis of Non-Inverting Amplifier Note that step 2 uses a voltage divider to find the voltage at VB relative to the output voltage. 2) + : VA = Vin Rg − : VB = Vout R f + Rg Rg 3) VA = VB Vin = Vout R f + Rg + V R + R 1) : out = f g − : Vin Rg V R out =1+ f Vin Rg The Voltage Follower V out =1 Vin analysis : 1] VA = Vout 2] VB = Vin VA = VB therefore, Vout = Vin Why is it useful? In this voltage divider, we get a different output depending upon the load we put on the circuit. Why? We can use a voltage follower to convert this real voltage source into an ideal voltage source. The power now comes from the +/- 15 volts to the op amp and the load will not affect the output. Part C Integrators and Differentiators General Op-Amp Analysis Differentiators Integrators Comparison Golden Rules of Op-Amp Analysis Rule 1: VA = VB • The output attempts to do whatever is necessary to make the voltage difference between the inputs zero. • The op-amp “looks” at its input terminals and swings its output terminal around so that the external feedback network brings the input differential to zero. Rule 2: IA = IB = 0 • The inputs draw no current • The inputs are connected to what is essentially an open circuit General Analysis Example(1) Assume we have the circuit above, where Zf and Zin represent any combination of resistors, capacitors and inductors. General Analysis Example(2) We remove the op amp from the circuit and write an equation for each input voltage. Note that the current through Zin and Zf is the same, because equation 1] is a series circuit. General Analysis Example(3) I Since I=V/Z, we can write the following: V −V V −V I = in A = A out Zin Z f But VA = VB = 0, therefore: Vin −V V Z = out out = − f Zin Z f Vin Zin General Analysis Conclusion For any op amp circuit where the positive input is grounded, as pictured above, the equation for the behavior is given by: V Z out = − f V Z in in Ideal Differentiator Phase shift jπ/2 - ±π Net-π/2 Amplitude changes by a factor of analysis : ωRfCin Vout Z f R f = − = − = − jω R f Cin Vin Zin 1 jω Cin Analysis in time domain I dV I = C Cin V = I R I = I = I Cin in dt Rf Rf f Cin Rf d(Vin −VA ) VA −Vout I = Cin = VA = VB = 0 dt R f dV therefore, V = −R C in out f in dt Problem with ideal differentiator Ideal Real Circuits will always have some kind of input resistance, even if it is just the 50 ohms or less from the function generator. Analysis of real differentiator I 1 Zin = Rin + jω Cin V Z R jω R C out = − f = − f = − f in Vin Zin 1 jω RinCin +1 Rin + jω Cin Low Frequencies High Frequencies V out Vout R f = − jω R f Cin = − Vin Vin Rin ideal differentiator inverting amplifier Comparison of ideal and non-ideal Both differentiate in sloped region.
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